# Solve. displaystyleintfrac{{{78}{x}}}{{{25}-{12}{x}+{4}{x}^{2}}}

Solve. $\int \frac{78x}{25-12x+4{x}^{2}}$
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SoosteethicU
The given integration is
$\int \frac{78x}{25-12x+4{x}^{2}}$
A simple calculation the integration becomes
$\int \frac{78x}{25-12x+4{x}^{2}}=7\int \frac{dx}{25-12x+4{x}^{2}}$
$=7\int \frac{dx}{4{x}^{2}-12x+25}$
$=7\int \frac{dx}{4\left({x}^{2}-3x+\frac{25}{4}\right)}$
$=\frac{7}{4}\int \frac{dx}{{x}^{2}-3x+\frac{25}{4}}$
$=\frac{7}{4}\int \frac{dx}{{x}^{2}-3x+\frac{25}{4}}$
$=\frac{7}{4}\int \frac{dx}{{x}^{2}-2\cdot x\cdot \frac{3}{2}+\frac{9}{4}+\frac{25}{4}-\frac{9}{4}}$
$=\frac{7}{4}\int \frac{dx}{{\left(x-\frac{3}{2}\right)}^{2}+\frac{16}{4}}$
$=\frac{7}{4}\int \frac{dx}{{\left(x-\frac{3}{2}\right)}^{2}+{\left(2\right)}^{2}}$
$=\frac{7}{4}×\frac{1}{2}{\mathrm{tan}}^{-1}\left(\frac{x-\frac{3}{2}}{2}\right)+C$

$=\frac{7}{8}{\mathrm{tan}}^{-1}\left(\frac{2x-3}{4}\right)+C$
Where C is the constant of integration.
Thus, the value of the integration is
$\int \frac{7dx}{25-12x+4{x}^{2}}=\frac{7}{8}{\mathrm{tan}}^{-1}\left(\frac{2x-3}{4}\right)+C$
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